Find Lengths of Sides of Triangle

Date: 04/20/2002 at 01:32:19
From: Bennette
Subject: Triangle geometry
Let ABC be a right-angled triangle with angle C = 90 degrees. Let
the bisectors of angle A and angle B intersect BC and CA at D and E
respectively. Given that CD=9 and CE=8, find the lengths of the
sides of ABC.
I thought of using similar triangles, but can't prove that triangle
BCE and ACD and ABC are similar.

Date: 04/22/2002 at 09:54:15
From: Doctor Floor
Subject: Re: Triangle geometry
Hi, Bennette,
Thanks for your question.
We use notation a = BC, b = AC and c = AB.
Note that CD:DB = b:c. This can be seen from the fact that the
distances from any point on AD to AB and AC are equal, as they are
from the center of the inscribed circle, which lies on AD. From that
the areas of triangles ADC and ADB are in the ratio of b and c
(having equal heights from b and c respectively), from which we see
that BD and DC must be in the same ratio, since on these bases the
altitude from A is equal.
So we have
CD 9 b
-- = --- = - ....[1],
DB a-9 c
and thus
(a-9)b = 9c
ab = 9b+9c........[2].
In a similar fashion we find
CE 8 a
-- = --- = -.....[3],
EA b-8 c
and thus
a(b-8)=8c
ab = 8a+8c.......[4].
Combining [2] and [4] gives
8a+8c = 9b+9c
c+9b
a = ---- .
8
We can substitute this into a^2 + b^2 = c^2, giving
c^2 + 18bc + 81b^2
------------------ + b^2 = c^2
64
c^2 + 18bc + 81b^2 + 64b^2 = 64c^2
145b^2 + 18bc - 63 c^2 = 0
145(b/c)^2 + 18(b/c) - 63 = 0.
From this quadratic equation we find that, since b/c must be positive,
b/c = 0.6. Substitution into [1] yields that a = 24. What's more, we
see that sin(<A) = b/c = 0.6, and this gives us without much trouble
that
cos(<A) = a/c = sqrt(1 - sin^2(<A)) = sqrt(1 - (0.6)^2) = 0.8.
And thus we find, knowing a, that c = 30 and consequently b = 18.
If you have more questions, just write back.
Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/